Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 2e5 + 5;
const int SZ = N * 25;
int n, m, q, x, y, z, val, sz, ch[SZ][2], S[SZ], rt[N], fir[N], ne[M], to[M],
    cnt, dep[N];
vector<int> T[N];
void add(int x, int y) {
  ne[++cnt] = fir[x];
  fir[x] = cnt;
  to[cnt] = y;
}
void link(int x, int y) {
  add(x, y);
  add(y, x);
}
void up(int x) {
  S[0] = 0;
  S[x] = S[ch[x][0]] + S[ch[x][1]];
}
void add(int &x, int l, int r, int val) {
  ++sz;
  ch[sz][0] = ch[x][0];
  ch[sz][1] = ch[x][1];
  S[sz] = S[x];
  x = sz;
  if (l == r) {
    S[x] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (val > mid)
    add(ch[x][1], mid + 1, r, val);
  else
    add(ch[x][0], l, mid, val);
  up(x);
}
int st[N][20];
void dfs(int x, int f) {
  dep[x] = dep[f] + 1;
  st[x][0] = f;
  for (int i = 1; i < 20; ++i) st[x][i] = st[st[x][i - 1]][i - 1];
  rt[x] = rt[f];
  for (int i = 0; i < (int)T[x].size(); ++i) add(rt[x], 1, m, T[x][i]);
  for (int i = fir[x]; i; i = ne[i]) {
    int V = to[i];
    if (V == f) continue;
    dfs(V, x);
  }
}
int LCA(int x, int y) {
  if (dep[x] > dep[y]) swap(x, y);
  for (int i = 19; i >= 0; --i)
    if (dep[st[y][i]] >= dep[x]) y = st[y][i];
  if (x == y) return x;
  for (int i = 19; i >= 0; --i)
    if (st[x][i] != st[y][i]) {
      x = st[x][i], y = st[y][i];
    }
  return st[x][0];
}
void query(int a, int b, int c, int d, int l, int r, int val) {
  if (l == r) {
    if (S[a] + S[b] - S[c] - S[d]) printf("%d ", l);
    return;
  }
  int mid = (l + r) >> 1;
  int All = S[ch[a][0]] + S[ch[b][0]] - S[ch[c][0]] - S[ch[d][0]];
  if (All >= val)
    return (void)query(ch[a][0], ch[b][0], ch[c][0], ch[d][0], l, mid, val);
  query(ch[a][0], ch[b][0], ch[c][0], ch[d][0], l, mid, All);
  query(ch[a][1], ch[b][1], ch[c][1], ch[d][1], mid + 1, r, val - All);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    link(x, y);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &x);
    T[x].push_back(i);
  }
  dfs(1, 0);
  for (; q--;) {
    scanf("%d%d%d", &x, &y, &z);
    int L = LCA(x, y);
    int now = st[L][0];
    int All = S[rt[x]] + S[rt[y]] - S[rt[L]] - S[rt[now]];
    z = min(z, All);
    printf("%d ", z);
    if (z) query(rt[x], rt[y], rt[L], rt[now], 1, m, z);
    puts("");
  }
}